Re: [中學] 三次根號
※ 引述《aaaasd ()》之銘言：
: 已知: (2^1/3-1)^1/3=a^1/3+b^1/3+c^1/3
: 求:a+b+c=?
: 非常感謝!
令 x^3 = 2
(x+1)^3 = x^3 + 3x^2 + 3x +1 = 3(x^2+x+1)
(x-1)(x+1)^3 = 3(x-1)(x^2+x+1) = 3(x^3-1) = 3
x-1 = 3 / (x+1)^3
(x-1)^(1/3) = 3^1/3 / x+1
= (3^1/3) * (x^2-x+1) / (x+1)(x^2-x+1)
= (3^1/3) * ( 2^2/3 - 2^1/3 +1 )/ x^3+1
= (3^1/3) * (2^2/3 - 2^1/3 + 1 )/ 3
= (2/3)^(-2/3) - 3^(-2/3) * 2^(1/3) + 3^(-2/3)
= (4/9)^(1/3) - (2/9)^(1/3) + (1/9)^(1/3)
a=4/9 b= -2/9 c=1/9
所以 a+b+c = (4-2+1)/9 = 1/3
希望有回答到你 QQ
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